MATHEMATICS MIDDLE SCHOOL

The half-life of carbon-14 is 5,730 years. How many grams of parent material will remain from a 120-gram sample of carbon-14 after 2 half-lives?90

60

30

15

Answers

Answer 1
Answer: t_{(1)/(2)}=5730\n N_0=120\ g\n t=2t_{(1)/(2)}=2\cdot5730=11460\n N(t)=?\n\n N(t)=N_0\left((1)/(2)\right)^{\frac{t}{t_{(1)/(2)}}}\n N(11460)=120\cdot\left((1)/(2)\right)^{(11460)/(5730)}\n N(11460)=120\cdot\left((1)/(2)\right)^2\n N(11460)=120\cdot(1)/(4)\n N(11460)=30\ g
Answer 2
Answer: After one half-life . . . 1/2 of the total parent material

After another half-life . . . 1/2 of the half remains = 1/4 of the total original parent

If the original parent is 120 grams, then (1/4 x 20) = 30 grams remains
after 2 half-lives.  It makes no difference what the substance is, or
how long its half-life is.

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Answers

Answer:

74

Step-by-step explanation:

you already answered youu ;LOLL
Answer to question: x = -4

Subtract. Reduce to lowest terms.
10 1/4 - 9 5/6

Answers

Answer:

5/12

Step-by-step explanation:

10 1/4 - 9 5/6

41/4 - 59/6

246/24 - 236/24 = 10/24 = 5/12

Answer:

-5, to simplify the expression

How do you graph y=-x+ 3/2 please help meee

Answers

I have attached the graph of y = -x + 1.5

The line goes through the points (0,1.5) and (1.5,0)

The base of a triangle is 2 more than twice the height of the triangle. Find the measure of the base and the height if the area is 20 ft^2 (the height is 4 and the base is 10, show a process so I can see how you got there

Answers

B = 2 + 2H 
A = 20 ft^2
H = H

Since the Area of a Triangle is Base * Height then the formula should be:
A = BH
20 = H(2+2H)
20 = 2H^2 + 2H
0 = 2H^2 + 2H - 20 

Find the zeros of the quadratic equation through substituting the values of a,b, and c. You will get the zeros of x = 2.7 and -3.7. The only thing that would make sense is you use the x=2.7 because there are no negative heights.

Since b = 2h+2 
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What is the value of y^2-x^4 when y=9 and x=2?

Answers

Answer:

65

Step-by-step explanation:

9^2 = 81

2^4 = 16

81-16=65

How many nanoseconds are in a year?

Answers


       (1 year) x (86,400 sec/day) x (365-1/4 day/yr) x (10⁹ nanos./sec)

  =  (1 x 86,400 x 365-1/4 x 10⁹) nanosec

  =    31,557,600 x 10⁹ nanosec

  =      3.15576 x 10¹⁶ nanosec

(This is a rough answer, because the 'year'
is not exactly  365-1/4 days.)

1 year = 3.1556926 × 10^16 nanoseconds       

hope this helps!!  

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